Answer:
108.9g of Silver can be produced from 125g of Ag2S
Explanation:
The compound Ag2S shows that two atoms of Silver Ag, combined with an atom of Sulphur S to form Ag2S. We can as well say the combination ration of Silver to Sulphur is 2:1
•Now we need to calculate the molecular weight of this compound by summing up the molar masses of each element in the compound.
•Molar mass of Silver Ag= 107.9g/mol
•Molar mass of Sulphur S= 32g/mol
•Molecular weight of Ag2S= (2×107.9g/mol) + 32g/mol
•Molecular weight of Ag2S= 215.8g/mol + 32g/mol= 247.8g/mol
•From our calculations, we know that 215.8g/mol of Ag is present in 247.8g/mol of Ag2S
If 247.8g Ag2S produced 215.8g Ag
125g Ag2S will produce xg Ag
cross multiplying we have
xg= 215.8g × 125g / 247.8g
xg= 26975g/247.8
xg= 108.85g
Therefore, 108.9g of Silver can be produced from 125g of Ag2S
Answer:
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<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in =
of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
The average rate of reaction over a given interval can be calculated by taking the difference of concentration on a particular given reactant, and dividing it by the total time. In this case, (1.00 M - 0.655 M)/30 s = 0.0115 M/s, or 0.0115 mol/L-s, and this is the final rate of reaction.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of
=
= 0.0458 M
Concentration of
=
= 0.0521 M
GIven that :
Ka = 
Thus; it's pKa = 4.72




pH ≅ 4.80